darkpool - 1 year ago 83

Python Question

I have the following pandas dataframe:

`column_01 column_02 value`

ccc aaa 1

bbb ddd 34

ddd aaa 98

I need to re-organise the dataframe such that

`column_01`

`column_01`

`column_02`

`column_01 column_02 value`

aaa ccc 1

bbb ddd 34

aaa ddd 98

I could obviously do this by iterating over the dataframe one row at a time, comparing

`column_01`

`column_02`

Is there a way to do this without iterating over every row individually?

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Answer Source

You can use:

```
df[['column_01','column_02']] =
df[['column_01','column_02']].apply(lambda x: sorted(x.values), axis=1)
print (df)
column_01 column_02 value
0 aaa ccc 1
1 bbb ddd 34
2 aaa ddd 98
```

Another solutions:

```
df[['column_01','column_02']] = pd.DataFrame(np.sort(df[['column_01','column_02']].values),
index=df.index, columns=['column_01','column_02'])
```

only with numpy array:

```
df[['column_01','column_02']] = np.sort(df[['column_01','column_02']].values)
print (df)
column_01 column_02 value
0 aaa ccc 1
1 bbb ddd 34
2 aaa ddd 98
```

Second solution is faster, because `apply`

use loops:

```
df = pd.concat([df]*1000).reset_index(drop=True)
In [177]: %timeit df[['column_01','column_02']] = pd.DataFrame(np.sort(df[['column_01','column_02']].values), index=df.index, columns=['column_01','column_02'])
1000 loops, best of 3: 1.36 ms per loop
In [182]: %timeit df[['column_01','column_02']] = np.sort(df[['column_01','column_02']].values)
1000 loops, best of 3: 1.54 ms per loop
In [178]: %timeit df[['column_01','column_02']] = (df[['column_01','column_02']].apply(lambda x: sorted(x.values), axis=1))
1 loop, best of 3: 291 ms per loop
```